In special relativity, time dilation is described by Δt' = γ Δt with γ = 1/√(1 - v^2/c^2). If a clock moves at speed v, what happens to the observed time interval Δt' compared with the proper time Δt?

In special relativity, time dilation is described by Δt' = γ Δt with γ = 1/√(1 - v^2/c^2). If a clock moves at speed v, what happens to the observed time interval Δt' compared with the proper time Δt?

• A. Δt' = Δt
• B. Δt' < Δt
• C. Δt' > Δt
• D. Δt' is unrelated to Δt

Answer: (C)

Time dilation shows that moving clocks run slower from the perspective of another observer. The time between ticks measured in the clock’s own rest frame is the proper time, Δt. In a frame where the clock is moving at speed v, the interval between those same two events is Δt' = γ Δt, with γ = 1/√(1 − v^2/c^2). Since γ > 1 for any nonzero v, Δt' is larger than Δt. So the observed interval between ticks is dilated (longer) compared with the clock’s own proper time. At small speeds, the difference is tiny, but as v approaches c, the dilation becomes significant.

Time dilation shows that moving clocks run slower from the perspective of another observer. The time between ticks measured in the clock’s own rest frame is the proper time, Δt. In a frame where the clock is moving at speed v, the interval between those same two events is Δt' = γ Δt, with γ = 1/√(1 − v^2/c^2). Since γ > 1 for any nonzero v, Δt' is larger than Δt. So the observed interval between ticks is dilated (longer) compared with the clock’s own proper time. At small speeds, the difference is tiny, but as v approaches c, the dilation becomes significant.